T(n,k)=Number of length n 0..k arrays with each partial sum starting from the beginning no more than sqrt(2) standard deviataionsdeviations from its mean

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R. H. Hardin, <a href="/A244903/b244903.txt">Table of n, a(n) for n = 1..9999</a>

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T(n,k)=Number of length n 0..k arrays with each partial sum starting from the beginning no more than sqrt(2) standard deviataions from its mean

2, 3, 4, 4, 7, 6, 5, 14, 21, 12, 4, 19, 52, 59, 24, 5, 22, 89, 198, 163, 42, 6, 33, 118, 431, 764, 447, 84, 7, 42, 199, 646, 2009, 2740, 1341, 168, 8, 57, 316, 1299, 3602, 9257, 10484, 3905, 312, 9, 68, 469, 2336, 8695, 20358, 45207, 40696, 11271, 624, 10, 87, 624

1,1

Table starts

...2.....3......4.......5........4.........5.........6..........7..........8

...4.....7.....14......19.......22........33........42.........57.........68

...6....21.....52......89......118.......199.......316........469........624

..12....59....198.....431......646......1299......2336.......3949.......5912

..24...163....764....2009.....3602......8695.....17392......33803......55282

..42...447...2740....9257....20358.....56609....130548.....282785.....518428

..84..1341..10484...45207...116272....382743...1019188....2438447....5021116

.168..3905..40696..215411...669516...2617609...7890336...21257775...48258938

.312.11271.159332.1061375..3880216..17614513..61161968..182555691..464345786

.624.32357.627156.5141999.22605350.117799671.475132844.1597877557.4477346412

Computation in integer form, using 6 times the 0..k mean and 36 times the variance, mean6(k)=3*k; var36(k)=6*k*(2*k+1)-mean6(k)^2; then (6*sum{x(i),i=1..j}-j*mean6(k))^2<=2*j*var36(k) for all j=1..n

Some solutions for n=6 k=4

..4....1....4....4....4....1....3....1....3....3....2....3....2....3....2....4

..0....1....1....1....1....4....2....4....1....3....3....1....1....3....0....1

..2....4....0....1....1....1....3....1....1....0....2....0....3....0....4....3

..2....2....0....0....4....0....0....4....4....2....0....3....1....3....4....2

..2....4....4....3....4....4....3....2....3....1....4....0....1....0....4....3

..3....0....0....3....2....1....3....1....0....3....4....2....1....3....1....3

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R. H. Hardin, Jul 07 2014

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